Linear operator $T$ whose square is the average of $T$ and $T^*$

Question

Having trouble with a question and I will be thankful for a hint.

Let $V$ be a an inner-product space with finite dimension over $\Bbb C$, $\space$ $T : V \to V$ linear transformation such that $T^2 =\frac12(T+T^*)$

A) Prove that $T$ is normal

B) Prove that $T^2-T=0$

So A was ok, since $T^*=2T^2-T$, and $T$ commutes with any polynomial in $T$, so $T$ is normal.

Having trouble with solving B, I've tried to use the fact that $T$ is diagonalizable and tried also some algebra techniques, but I feel like I'm missing something.

Sorry for my horrible English, I would appreciate any help.

Thanks!

Answer 1

HINT: If $T$ is normal and $v$ is an eigenvector with eigenvalue $\lambda$, then $v$ is also an eigenvector of $T^*$ with eigenvalue $\overline\lambda$. From this you conclude that any eigenvalue $\lambda$ must satisfy $\overline\lambda = 2\lambda^2-\lambda$. Do some algebra to conclude that $\lambda$ must be either $0$ or $1$, and so $T=T^*$ and $T$ is a projection.

Answer 2

Hint: applying the given equation to any eigenvector of eigenvalue $\lambda\in\Bbb C$, gives that the eigenvalue satisfies $\lambda^2=\frac12(\lambda+\overline\lambda)=\Re\lambda$. It is easy to see that the only solutions to that equation are real, and then $\lambda^2-\lambda=0$ follows.

And a diagonalisable operator, all of whose eigenvalues are roots of a given polynomial (here $X^2-X$), is itself annihilated by that polynomial.